Solution of SEAMO 2017 Paper C
SEAMO 2017 Paper C No. 1
A 3-digit number is such that it is equal to 19 times the sum of its digits. What is its largest possible value?A. 114
B. 133
C. 152
D. 399
E. None of the above
Solution: Show/Hide
$\overline{abc}=19(a+b+c)$$100a+10b+c=19a+19b+19c$
$81a-9b-18c=0$
$9(9a-b-2c)=0$
$9a-b-2c$ must be equal to 0.
For $a=3$
$\begin{align}9(3)-b-2c &= 0 \\ b &= 27-2c \end{align}$
For $b\le 9$,
we need
$\begin{align}27-2c &\le 9 \\ -2c &\le 9-27 \\ -2c &\le -18 \\ c &\ge 9 \end{align}$
Let $c=9$
$\begin{align}b &= 27-2c \\ &= 27-2(9) \\ b &= 9 \end{align}$
So, the number is N = $\overline{abc}=399$
Answer: D
SEAMO 2017 Paper C No. 2
A rope 580 cm long is to be cut into 40 cm and 90 cm segments without any wastage. How many ways are there to do this?A. 1
B. 2
C. 3
D. 4
E. 5
Solution: Show/Hide
$40x+90y=580$The following values satisfy
$x=10$, $y=2$
$x=1$, $y=6$
Thus, 2 ways.
Answer: B
SEAMO 2017 Paper C No. 3
A new operation is defined as$2\oplus 4=2+3+4+5=14$
$5\oplus 3=5+6+7=18$
Find the value of $m$ in $m\oplus 7=49$
A. 1
B. 2
C. 3
D. 4
E. 5
Solution: Show/Hide
$m\oplus 7$ = 49$m+(m+1)+(m+2)+..+(m+6)$ = 49
$7m+1+2+3+4+5+6$ = 49
$7m+21$ = 49
$7m = 49-21$
$7m = 28$
$m = 4
Answer: D
SEAMO 2017 Paper C No. 4
A car travelled at 40 km/h for the first 2 hours. It travelled at 60 km/h for the last 3 hours. What was its average speed?A. 50
B. 51
C. 52
D. 54
E. None of the above
Solution: Show/Hide
Average speed = $\frac{\text{Total distance}}{\text{Total time}}$= $\frac{40\times 2+60\times 3}{2+3}$
= $\frac{260}{5}$
= 52 km/h
Common mistake: $\frac{40+60}{2}$ = 50 km/h, which is wrong.
Answer: C
SEAMO 2017 Paper C No. 5
Marks fills in each circle a number from 1, 2, 3, …, 8, such that the sum of numbers at all corners of any triangle is 12.Find (a + b + c + d).
A. 10
B. 11
C. 12
D. 13
E. 14
Solution: Show/Hide
Note: $a$, $b$, $c$ and $d$ are repeated when all triangles are added.$12\times 4=48$
1 + 2 + 3 + … + 7 = 36
48 – 36 = 12
Thus, $a+b+c+d=12$
Answer: C
SEAMO 2017 Paper C No. 6
Lines AC and BD divide the quadrilateral ABCD into 4 triangles of different areas. Given that BE : DE = 2 : 1 and AE : CE = 1 : 3, find the ratio of the areas $\Delta ADE:\Delta BCE$.
A. 3 : 7
B. 2 : 5
C. 1 : 3
D. 1 : 4
E. None of the above
Solution: Show/Hide
Since $\Delta ABE$ and $\Delta ADE$ share the same height.
Ratio of areas $\Delta ADE:\Delta ABE=1:2$
Since $\Delta ABE$ and $\Delta BCE$ share the same height.
Ratio of areas $\Delta ABE:\Delta BCE=1:3=2:6$
Thus, ratio of areas $\Delta ADE:\Delta BCE:\Delta BCE=1:2:6$
Answer: E
SEAMO 2017 Paper C No. 7
Find the value of 1 – 2 + 3 – 4 + … + 2015 – 2016 + 2017
A. 1006
B. 1007
C. 1008
D. 1009
E. None of the above
Solution: Show/Hide
1 – 2 + 3 – 4 + … + 2015 – 2016 + 2017= (1 – 2) + (3 – 4) + … + (2015 – 2016) + 2017
= $\underbrace{(-1)+(-1)+...+(-1)}_{\text{(-1) there are as many as 1008}}+2017$
= $1008\times (-1)+2017$
= 1009
Answer: D
SEAMO 2017 Paper C No. 8
Melvin used the number 1, 2, 3, 4, 5, 6 and 7, without repeating, to form three 2-digit numbers and one 1-digit number. The sum of the four numbers is 100. Find the largest 2-digit number Melvin formed.A. 54
B. 57
C. 61
D. 63
E. None of the above
Solution: Show/Hide
Let the 4 numbers be $\overline{ab}$, $\overline{cd}$, $\overline{ef}$, $g$$\overline{ab}=10a+b$
$\overline{cd}=10c+d$
$\overline{ef}=10e+f$
We have $10(a+c+e)+(b+d+f+g)=100$
We know that $(b+d+f+g)$ must be a multiple of 10, either 10 or 20
Since 1 + 2 + 3 + … + 7 = 28
When $(b+d+f+g)=10$,
$(a+c+e)=28-10=18$
$10(a+c+e)+(b+d+f+g)=190$ (reject)
When $(b+d+f+g)=20$
$(a+c+e)=8$
$10(a+c+e)+(b+d+f+g)=100$ (reject)
Try: $b$, $d$, $f$, $g$ $\to $ 3, 4, 6, 7
$a$, $c$, $e$ $\to $ 1, 2, 3
$\overline{ef}=57$
Answer: B
SEAMO 2017 Paper C No. 9
$<n>$ denotes the smallest whole number that is not a factor of $n$.For example,
$<7>$ = 2
$<12>$ = 5
Find the value of $<<23>\times <100>>$
A. 4
B. 5
C. 6
D. 8
E. 10
Solution: Show/Hide
The factors of 23 are 1 and 23, so $<23>$ = 2.The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100
The smallest whole number is 1, and 2 is a factor as well.
The next whole number is 3, so $<100>$ = 3
$<<23>\times <100>>$
= $<2\times 3>$
= $<6>$, the factors of 6 are 1, 2, 3, 6
= 4
Answer: A
SEAMO 2017 Paper C No. 10
When a 6-digit number $\overline{abcdef}$ is multiplied by 3, the result is $\overline{bcdef9}$.
Find the value of $\overline{abcdef}$.
A. 286713
B. 284713
C. 297813
D. 279813
E. None of the above
Solution: Show/Hide
$f=3$ as $3\times 3=9$$\begin{align}3\times e &= f \\ 3\times e &= 3 \\ e &= 1 \end{align}$
$d=7$ as $3\times 7=21$
$c=5$ as $3\times 5+2=17$
$b=8$ as $3\times 8+1=25$
$a=2$ as $3\times 2+2=8$
$\overline{abcdef}$ = 285713
None of the above.
Answer: E
SEAMO 2017 Paper C No. 11
Find the remainder when the ${{2017}^{th}}$ term of the following sequence is divided by 5.1, 2, 4, 7, 11, 16, 22, 29, …
A. 1
B. 2
C. 3
D. 4
E. 0
Solution: Show/Hide
Observe that pattern of remainders:
The pattern recurs: 1, 2, 4, 2, 1, ….
$2017\div 5$ = 403 R 2
The second term is 2.
Answer: B
SEAMO 2017 Paper C No. 12
All digits are different in the electronic clock display below.
How many times between 7:00:00 and 8:00:00 will all digits be different?
A. 1242
B. 1248
C. 1254
D. 1260
E. None of the above
Solution: Show/Hide
Let the format of digital clock be$h:m_1m_2:s_1s_2$
$h$ only one choice, namely 7
$m_1$ and $s_1$ each to be filled in with digits 0 to 5, 6 choices for $m_1$ and 5 choices for $s_1$ (different from $m_1$)
$m_2$ and $s_2$ each to be filled in with digits 0 to 9, but not 7 and not the numbers used in $m_1$ and $s_1$, so 7 choices for $m_2$ and 6 choices for $s_2$ (different from $m_2$)
$1\times 6\times 7\times 5\times 6=1260$
Answer: D
SEAMO 2017 Paper C No. 13
How many numbers are the same in both number patterns below?1, 3, 5, 7, 9, …, 2015, 2017
1, 4, 7, 10, …, 2014, 2017
A. 337
B. 338
C. 339
D. 340
E. None of the above
Solution: Show/Hide
Differences are 2 and 3 in $1^{st}$ and $2^{nd}$ sequences respectively.We are looking for 1, 7, 13, 19, 25, … in both.
Difference is 6 here, ($2\times 3=6$).
$(2017-1)\div 6=336$
336 + 1 = 337
Answer: A
SEAMO 2017 Paper C No. 14
The figure shows a semicircle and a sector overlapping each other. Find the perimeter of the shaded region. Take $\pi =\frac{22}{7}$.
A. 43
B. $43\frac{1}{3}$
C. $43\frac{2}{3}$
D. 44
E. None of the above
Solution: Show/Hide
14 + $\frac{1}{2}\times \pi \times d$ + $\frac{30{}^\circ }{360{}^\circ }\times 2\times \pi \times r$= 14 + $\frac{1}{2}\times \frac{22}{7}\times 14$ + $\frac{1}{12}\times 2\times \frac{22}{7}\times 14$
= 14 + 22 + $\frac{22}{3}$
= $43\frac{1}{3}$ cm
Answer: B
SEAMO 2017 Paper C No. 15
A camera was priced at 35% more than its original price and later sold at a 10% discount. The buyer also redeemed a $\$$50 cash voucer. The net profit from the sale was $\$$380.Find the original price of the camera.
A. $1872
B. $1900
C. $2000
D. $2172
E. None of the above
Solution: Show/Hide
Let the cost price be C.$\begin{align}0.9\times (100\%+35\%)C-50 &= 380+C \\ \frac{1215}{1000}C-C &= 430 \\ \frac{215}{1000}C &= 430 \\ 215C &= 430000 \\ C &= 2000 \end{align}$
The original price of the camera is $\$$2000.
Answer: C
SEAMO 2017 Paper C No. 16
Find the largest and smallest fractions in $\frac{9}{5}$, $\frac{12}{7}$, $\frac{27}{17}$, $\frac{36}{19}$, $\frac{54}{29}$
A. $\frac{54}{29}$, $\frac{9}{5}$
B. $\frac{12}{7}$, $\frac{27}{17}$
C. $\frac{36}{19}$, $\frac{54}{29}$
D. $\frac{12}{7}$, $\frac{27}{17}$
E. $\frac{36}{19}$, $\frac{27}{17}$
Solution: Show/Hide
$\frac{9}{5}$, $\frac{12}{7}$, $\frac{27}{17}$, $\frac{36}{19}$, $\frac{54}{29}$$1\frac{4}{5}$, $1\frac{5}{7}$, $1\frac{10}{17}$, $1\frac{17}{19}$, $1\frac{27}{29}$
The largest is $1\frac{4}{5}=\frac{9}{5}$ and smallest is $1\frac{10}{17}=\frac{27}{17}$.
Answer: E
SEAMO 2017 Paper C No. 17
Alan and Ben are working on a science project. Together, they can finish it in 35 days. If Alan works alone for 40 days, Ben will take a further 28 days to complete the project. If Alan works alone for 30 days, how many more days will Ben take to complete the project?A. 40
B. 42
C. 38
D. 44
E. None of the above
Solution: Show/Hide
40 days by A + 28 days by B = 1 project35 days by A + 35 days by B = 1 project
Solve above 2 equation get ration A : N = 7 : 5
7 + 5 = 12
B: 35 × 12/5 = 84 days alone, so in one day B does 1/84 work.
A: 35 × 12 / 7 = 60 days alone, so in one day A does 1/60 work.
A: 1/60 × 30 days = ½ of wordk.
So ½ of work left for B.
B = 1/84 × number of days = ½,
so number of days = ½ × 84 = 42 days.
Answer: B
SEAMO 2017 Paper C No. 18
Evaluate$\left( \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017} \right)\left( 1+\frac{1}{2}+...+\frac{1}{2016} \right)$ $-$ $\left( 1+\frac{1}{2}+...+\frac{1}{2017} \right)\left( \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016} \right)$
A. $\frac{1}{2015}$
B. $\frac{1}{2016}$
C. $\frac{1}{2017}$
D. $\frac{1}{2013}$
E. None of the above
Solution: Show/Hide
Let $A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}$ and $B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}$$\left( \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017} \right)\left( 1+\frac{1}{2}+...+\frac{1}{2016} \right)-\left( 1+\frac{1}{2}+...+\frac{1}{2017} \right)\left( \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016} \right)$
= $A\left( 1+B \right)-\left( 1+A \right)B$
= $A+AB-B-AB$
= $A-B$
$\frac{\begin{align}A &= \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017} \\ B &= \frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016} \\ \end{align}}{A-B=\frac{1}{2017}}-$
Answer: C
SEAMO 2017 Paper C No. 19
A mall escalator moves upwards at a constant rate from the $1^{st}$ to $3^{rd}$ floor.Chiranjit walks up the escalator at a constant rate of 2 steps per second, and arrives at the third floor after 40 seconds.
Debdeep walks up the escalator at a constant rate of 3 steps every 2 second, and arrives at the third floor after 50 seconds.
How many steps are there on the escalator?
A. 100
B. 90
C. 80
D. 70
E. 60
Solution: Show/Hide
Let the speed of escalator be $e$$\begin{align}40\times 2+40e &= 50\times \frac{3}{2}+50e \\ 80+40e &= 75+50e \\ 40e-50e &= 75-80 \\ -10e &= -5 \\ e &= -5\div (-10) \\ e &= \frac{1}{2}\,\text{step/s} \end{align}$
$40\times 2+40e$ = $40\times 2+40\times \frac{1}{2}$ = 100 steps.
or
$50\times \frac{3}{2}+50e$ = $50\times \frac{3}{2}+50\times \frac{1}{2}$ = 100 steps.
Answer: A
SEAMO 2017 Paper C No. 20
Three identical circles of radii 7 cm intersect through their centers as shown in the figure. Find the area of the shaded region. Take $\pi =\frac{22}{7}$.
A. 22
B. 66
C. $66\frac{2}{7}$
D. 77
E. None of the above
Solution: Show/Hide
By cutting and pasting, we obtain a figure of a semicircle.
$\frac{1}{2}\times \frac{22}{7}\times 7\times 7=11\times 7=77\,\text{cm}^2$
Answer: 77
SEAMO 2017 Paper C No. 21
Evaluate$\frac{(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)}{2^{32}-1}$
Solution: Show/Hide
$\frac{(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)}{2^{32}-1}$= $\frac{(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)}{(2^{16}-1)(2^{16}+1)}$
= $\frac{(2+1)(2^2+1)(2^4+1)(2^8+1)}{(2^8-1)(2^8+1)}$
= $\frac{(2+1)(2^2+1)(2^4+1)}{(2^4-1)(2^4+1)}$
= $\frac{(2+1)(2^2+1)}{(2^2-1)(2^2+1)}$
= $\frac{(2+1)}{(2-1)(2+1)}$
= 1
Answer: 1
SEAMO 2017 Paper C No. 22
Find the value of$1^3+2^3+3^3+4^3+...+19^3+20^3$
Solution: Show/Hide
Remember:$1^3+2^3+3^3+...+n^3$ = $(1+2+3+..+n)^2$
$1^3+2^3+3^3+...+20^3$
= $(1+2+3+..+20)^2$
= $\left( \frac{20(20+1)}{2} \right)^2$
= $(210)^2 = 44100
Answer: 44100
SEAMO 2017 Paper C No. 23
2017 soldiers were lined up in a single file. The commander ordered them to number off 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, … starting from the first soldier. He then ordered them to number off 1, 2, 3, 4, 1, 2, 3, 4, 1, … starting from the last soldier. How many soldiers had the same number both times?Solution: Show/Hide
Find soldiers:
2017 + 5 = 403 R2
2017 + 4 = 504 R1
Every 20 soldiers there are 4 with the same numbers.
2017 + 20 = 100 R17
So, 100 × 4 = 400, but there are 4 more soldiers with the same numbers amongst the last 17 soldiers.
So, 404 + 4 = 404.
Answer: 404
SEAMO 2017 Paper C No. 24
Given that $S=\frac{1}{\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+...+\frac{1}{2017}}$.Find the largest whole number smaller than S.
Solution: Show/Hide
Assume all to be $\frac{1}{2008}$,$\frac{1}{\frac{1}{2008}+\frac{1}{2008}+\frac{1}{2008}+...+\frac{1}{2008}}$
= $\frac{1}{\frac{10}{2008}}$
= $\frac{2008}{10}$
= 200.8
Assume all to be $\frac{1}{2017}$,
$\frac{1}{\frac{1}{2017}+\frac{1}{2017}+\frac{1}{2017}+...+\frac{1}{2017}}$
= $\frac{1}{\frac{10}{2017}}$
= $\frac{2017}{10}$
= 201.7
$200.8<S<201.7$
Largest whole number is 201.
Answer: 201
SEAMO 2017 Paper C No. 25
Find the area of the shaded regions in the figure, given that the area of square is 1 $\text{cm}^2$. Take $\pi $ as $\frac{22}{7}$.
Solution: Show/Hide
The 4 radii are 2 cm, 3 cm, 4 cm, 5 cmArea shaded:
$\frac{1}{4}\pi \times 2^2$ + $\frac{1}{4}\pi \times 3^2$ + $\frac{1}{4}\pi \times 4^2$ + $\frac{1}{4}\pi \times 5^2$
= $\frac{1}{4}\pi (2^2+3^2+4^2+5^2)$
= $\frac{1}{4}\pi (4+9+16+25)$
= $\frac{1}{4}\times \frac{22}{7}\times 54$
= $\frac{297}{7}\,\text{cm}^2$
Answer: 297/7
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